错误！ 'float' 和 'const char [2]' 类型的无效操作数到二进制 'operator<<' [关闭]答案

【问题标题】：ERROR! invalid operands of types 'float' and 'const char [2]' to binary 'operator<<' [closed]错误！ 'float' 和 'const char [2]' 类型的无效操作数到二进制 'operator<<' [关闭]
【发布时间】：2020-11-28 17:45:14
【问题描述】：

我正在尝试制作一个相当于两个分数的四函数计算器的程序。错误出现在 cout 语句中的每个 switch 案例中。错误是：

[错误] 'float' 和 'const char [2]' 类型的无效操作数二进制'运算符

我尝试谷歌搜索，没有找到，但我确实发现了相同的错误，但使用 int 而不是 float。不幸的是，他犯了一个明显的错误，他甚至没有使用 cout 语句并将

#include <iostream>
#include <conio.h>

using namespace std;

main()
{   //This Program is an equivalent of a four function calculator for fraction
    float N1, D1, N2, D2;
    char OP, slash;
    system("cls");
    
    cout<<"Hint: Enter Both Fractions with Operator in between; 1/2 + 6/3";
    cout<<"\nEnter Fractionaly Binary Operation: ";
    cin>>N1>>slash>>D1>>OP>>N2>>slash>>D2;

    if(slash=='/')
    {
        switch(OP) 
        {
            case '+':
                cout<<"Addition of Given Fraction is: ("<<N1<<"/"<<D1<<") + ("<<N2<<"/"<<D2<<") = "<<(((N1*D2)+(N2*D1))<<"/"<<(D1*D2))<<" = "<<(((N1*D2)+(N2*D1))/(D1*D2))<<endl;
                break;

            case '-':
                cout<<"Subtraction of Given Fraction is: ("<<N1<<"/"<<D1<<") - ("<<N2<<"/"<<D2<<") = "<<(((N1*D2)-(N2*D1))<<"/"<<(D1*D2))<<" = "<<(((N1*D2)-(N2*D1))/(D1*D2))<<endl;
                break;

            case '*':
                cout<<"Multiplication of Given Fraction is: ("<<N1<<"/"<<D1<<") * ("<<N2<<"/"<<D2<<") = "<<((N1*N2)<<"/"<<(D1*D2))<<" = "<<((N1*N2)/(D1*D2))<<endl;
                break;

            case '/':
                cout<<"Division of Given Fraction is: ("<<N1<<"/"<<D1<<") / ("<<N2<<"/"<<D2<<") = "<<((N1*D2)<<"/"<<(D1*N2))<<" = "<<((N1*D2)/(D1*N2))<<endl;
                break;

            default:
                cout<<"INVALID OPERATOR!";
                getch();
                system("cls");
                return(0);
        }
    }
    else
    {
        cout<<"INVALID INPUT!";
        getch();
        system("cls");
        return(0);
    }
    
    
    getch();
    system("cls");
    return(0);
}

同样的错误出现在以下几行中..

                case '+':
                    cout<<"Addition of Given Fraction is: ("<<N1<<"/"<<D1<<") + ("<<N2<<"/"<<D2<<") = "<<(((N1*D2)+(N2*D1))<<"/"<<(D1*D2))<<" = "<<(((N1*D2)+(N2*D1))/(D1*D2))<<endl;
                    break;
    
                case '-':
                    cout<<"Subtraction of Given Fraction is: ("<<N1<<"/"<<D1<<") - ("<<N2<<"/"<<D2<<") = "<<(((N1*D2)-(N2*D1))<<"/"<<(D1*D2))<<" = "<<(((N1*D2)-(N2*D1))/(D1*D2))<<endl;
                    break;
    
                case '*':
                    cout<<"Multiplication of Given Fraction is: ("<<N1<<"/"<<D1<<") * ("<<N2<<"/"<<D2<<") = "<<((N1*N2)<<"/"<<(D1*D2))<<" = "<<((N1*N2)/(D1*D2))<<endl;
                    break;
    
                case '/':
                    cout<<"Division of Given Fraction is: ("<<N1<<"/"<<D1<<") / ("<<N2<<"/"<<D2<<") = "<<((N1*D2)<<"/"<<(D1*N2))<<" = "<<((N1*D2)/(D1*N2))<<endl;

我必须将此作为作业提交。

这是我想要的界面。注意：这是使用 cout 语句进行的，后端没有计算。

【问题讨论】：

你只是有很多不匹配的括号。尝试将中间值存储在变量中以提高代码清晰度。
main 在 C++ 中的返回类型必须为 int。
我检查了所有括号，它们是匹配的。似乎没有任何不匹配的。关于int，所有其他程序都可以在没有它的情况下工作吗？不管怎样，让我试试吧..
即使使用 int main 仍然给出相同的错误
@FrançoisAndrieux 我为所需界面添加了一个模型。看看吧，也许你会比我更了解这个程序。 :D

标签： c++ dev-c++

【解决方案1】：

检查括号：

cout<<"Addition of Given Fraction is: ("<<N1<<"/"<<D1<<") + ("<<N2<<"/"<<D2<<") = "
<<(((N1*D2)+(N2*D1))<<"/"<<(D1*D2))<<" = "<<(((N1*D2)+(N2*D1))/(D1*D2))<<endl;
  ^                               ^

试试这个：

    case '+':
        cout<<"Addition of Given Fraction is: ("<<N1<<"/"<<D1<<") + ("<<N2<<"/"<<D2<<") = "<<((N1*D2)+(N2*D1))<<"/"<<(D1*D2)<<" = "<<(((N1*D2)+(N2*D1))/(D1*D2))<<endl;
        break;

    case '-':
        cout<<"Subtraction of Given Fraction is: ("<<N1<<"/"<<D1<<") - ("<<N2<<"/"<<D2<<") = "<<((N1*D2)-(N2*D1))<<"/"<<(D1*D2)<<" = "<<(((N1*D2)-(N2*D1))/(D1*D2))<<endl;
        break;

    case '*':
        cout<<"Multiplication of Given Fraction is: ("<<N1<<"/"<<D1<<") * ("<<N2<<"/"<<D2<<") = "<<(N1*N2)<<"/"<<(D1*D2)<<" = "<<((N1*N2)/(D1*D2))<<endl;
        break;

    case '/':
        cout<<"Division of Given Fraction is: ("<<N1<<"/"<<D1<<") / ("<<N2<<"/"<<D2<<") = "<<(N1*D2)<<"/"<<(D1*N2)<<" = "<<((N1*D2)/(D1*N2))<<endl;
        break;

【讨论】：

【解决方案2】：

括号不匹配

cout<<"Addition of Given Fraction is: ("<<N1<<"/"<<D1<<") + ("<<N2<<"/"<<D2<<") = "<<

(((N1*D2)+(N2*D1))<<"/"<<(D1*D2))   /*<-here*/

<<" = "<<(((N1*D2)+(N2*D1))/(D1*D2))<<endl;

(((N1*D2)+(N2*D1))<<"/"<<(D1*D2))在一个街区内。

【讨论】：